Sequences, growth and decay (QCE General Mathematics Unit 3)

Syllabus dot point

“Use recursion to generate arithmetic and geometric sequences, display their terms in tabular and graphical form, use the rules t(n) = t(1) + (n - 1)d and t(n) = t(1)r^(n - 1), and model and analyse practical situations involving linear and geometric growth and decay, such as simple interest, taxi fares and population growth (use of logarithms not required)”

15 min readUpdated 23 Sep 2026

Quick answer

An arithmetic sequence adds a constant $d$ ( $t_{n+1} = t_n + d$ , $t_n = t_1 + (n - 1)d$ ) and models linear growth or decay, such as simple interest, taxi fares and straight-line depreciation. A geometric sequence multiplies by a constant ratio $r$ ( $t_{n+1} = rt_n$ , $t_n = t_1r^{n-1}$ ) and models exponential growth or decay, such as doubling bacteria and diminishing-value depreciation, with $r = 1 \pm \text{rate}$ . Plot the terms as separate points, watch which term represents "after $k$ periods", and find when a target is reached by trial and improvement.

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What this dot point is asking

A sequence is an ordered list of numbers, called terms: $t_1, t_2, t_3, \ldots$ . Many real quantities change in regular steps: a taxi fare goes up by the same amount every kilometre, and a bacterial population doubles every hour. The QCAA General Mathematics 2025 syllabus (Unit 3, Topic 4, Growth and decay in sequences) asks you, for both arithmetic and geometric sequences, to

use recursion to generate the sequence,
display its terms in tabular and graphical form, and show that arithmetic sequences model linear growth and decay, and geometric sequences model exponential (geometric) growth and decay, in discrete situations,
use the rule for the $n$ th term: $t_n = t_1 + (n - 1)d$ (arithmetic) and $t_n = t_1r^{n-1}$ (geometric), and
model and analyse practical situations: simple interest, taxi fares, straight-line depreciation (arithmetic); population growth such as bacteria doubling each hour, and diminishing-value depreciation (geometric).

The syllabus states that logarithms are not required, so "how long until" questions are answered by trial and improvement or a table of values. Depreciation has its own page.

The answer

Notation

$t_n$ is the $n$ th term, $t_1$ is the first term and $n$ is the term number (1, 2, 3, ...). If $t_1$ is the value now, then the value after $k$ periods is $t_{k+1}$ . Watch this off-by-one carefully in every question.

Arithmetic sequences

An arithmetic sequence adds the same amount, the common difference $d$ , to get each new term:

t_{n+1} = t_n + d \qquad \text{(recursion, with } t_1 \text{ given)}

t_n = t_1 + (n - 1)d \qquad \text{(rule for the } n\text{th term)}.

The rule makes sense: to reach $t_n$ from $t_1$ you add $d$ exactly $n - 1$ times.

$d > 0$ : linear growth. $d < 0$ : linear decay.
To test a table: are the differences $t_2 - t_1$ , $t_3 - t_2$ , ... all equal?
Graph: the points lie on a straight line with gradient $d$ .

Contexts. Simple interest (the same interest every year), taxi fares (flag fall plus a fixed charge per kilometre), straight-line depreciation (the same loss every year), seats in rows that increase by the same number each row.

Geometric sequences

A geometric sequence multiplies by the same number, the common ratio $r$ , to get each new term:

t_{n+1} = r\,t_n \qquad \text{(recursion, with } t_1 \text{ given)}

t_n = t_1 r^{n-1} \qquad \text{(rule for the } n\text{th term)}.

$r > 1$ : geometric (exponential) growth. $0 < r < 1$ : geometric decay.
A growth rate of $p\%$ per period gives $r = 1 + \frac{p}{100}$ ; a decay rate of $p\%$ gives $r = 1 - \frac{p}{100}$ . For 4% growth, $r = 1.04$ ; for 30% decay, $r = 0.7$ .
To test a table: are the ratios $\frac{t_2}{t_1}$ , $\frac{t_3}{t_2}$ , ... all equal?
Graph: the points lie on a curve that gets steeper (growth) or flattens towards 0 (decay), but a decaying geometric sequence never reaches 0.

Contexts. Populations and bacteria (doubling means $r = 2$ ), compound interest, diminishing-value depreciation, medicine clearing from the body, a ball bouncing to a fixed fraction of its previous height.

Tables and graphs

Because $n$ only takes whole-number values, these are discrete models: plot separate points, not a joined line. The graph below compares two salary plans that both start at $50 000. Plan A adds $2500 a year (arithmetic: the points are in a straight line). Plan B grows by 4% a year (geometric: the points curve upwards). Plan B starts behind but overtakes A in year 13.

Year $n$	1	2	3	10	12	13	16
Plan A: $50\,000 + 2500(n-1)$	50 000	52 500	55 000	72 500	77 500	80 000	87 500
Plan B: $50\,000 \times 1.04^{n-1}$	50 000	52 000	54 080	71 165.59	76 972.70	80 051.61	90 047.18

Finding when a target is reached (no logarithms)

To find the first term that passes a value, use the rule and trial and improvement, or list terms with the calculator. Show the two terms either side of the target:

Arithmetic can also be solved with an inequality: $18 + 3(n - 1) > 80$ gives $n - 1 > 20.67$ , so $n = 22$ .
Geometric: 500 bacteria doubling each hour first exceed 1 000 000 when $500 \times 2^{k} > 1\,000\,000$ . $2^{10} = 1024$ gives 512 000 and $2^{11} = 2048$ gives 1 024 000, so after 11 hours.

Finding the rule from two terms

Arithmetic. If $t_3 = 17$ and $t_8 = 42$ , there are 5 steps from term 3 to term 8, so $5d = 25$ , $d = 5$ , and $t_1 = 17 - 2(5) = 7$ .
Geometric. If $t_2 = 12$ and $t_4 = 48$ , there are 2 steps, so $r^2 = 4$ and $r = 2$ (for positive terms), and $t_1 = 12 \div 2 = 6$ .

Key fact

Arithmetic: $t_{n+1} = t_n + d$ , $t_n = t_1 + (n - 1)d$ ; constant difference; linear growth or decay; points in a straight line. Geometric: $t_{n+1} = rt_n$ , $t_n = t_1r^{n-1}$ ; constant ratio; $r = 1 + \text{rate}$ for growth and $r = 1 - \text{rate}$ for decay; points on a curve. If $t_1$ is the starting value, the value after $k$ periods is $t_{k+1}$ . Use trial and improvement, not logarithms, to find when a target is reached.

Worked examples: arithmetic and geometric models

1. Arithmetic, geometric or neither?

Classify: (a) 5, 9, 13, 17; (b) 3, 6, 12, 24; (c) 2, 5, 10, 17.

(a) Differences 4, 4, 4: arithmetic, $d = 4$ .

(b) Ratios $\frac{6}{3} = \frac{12}{6} = \frac{24}{12} = 2$ : geometric, $r = 2$ .

(c) Differences 3, 5, 7 (not constant); ratios 2.5, 2, 1.7 (not constant): neither.

2. A taxi fare

A taxi charges $6.30 for a 1 km trip, and each extra kilometre adds $2.10. Find the fare for an 18 km trip and the longest whole-kilometre trip for $50.

Model: Arithmetic: $t_1 = 6.30$ , $d = 2.10$ , so $t_n = 6.30 + 2.10(n - 1)$ , where $n$ is the trip length in km.
18 km: $t_{18} = 6.30 + 2.10 \times 17 = 6.30 + 35.70 = 42.00$ , so $42.00.
$50 limit: $t_{21} = 6.30 + 2.10 \times 20 = 48.30$ and $t_{22} = 50.40$ . The longest trip is 21 km.

Marker's note: the flag fall is built into $t_1$ . An equivalent rule is fare $= 4.20 + 2.10n$ ; both give the same answers.

3. Simple interest

$5000 is invested at 4.5% p.a. simple interest. Model the balance and find when it first exceeds $7000.

Model. The interest is the same every year: $0.045 \times 5000 = 225$ . With $t_n$ the balance at the end of year $n$ : $t_1 = 5225$ , $d = 225$ , $t_n = 5225 + 225(n - 1)$ .

Target. $t_8 = 6800$ and $t_9 = 7025$ , so the balance first exceeds $7000 at the end of year 9.

Marker's note: under compound interest the interest grows each year, so the model would be geometric instead.

4. Bacterial growth

A culture starts with 500 bacteria and doubles every hour. How many are there after 6 hours, and when does the count first exceed one million?

Model: Geometric: $t_1 = 500$ (now, 0 hours), $r = 2$ , $t_n = 500 \times 2^{n-1}$ .
After 6 hours: That is $t_7 = 500 \times 2^6 = 32\,000$ bacteria.
One million: $t_{11} = 500 \times 2^{10} = 512\,000$ and $t_{12} = 500 \times 2^{11} = 1\,024\,000$ . $t_{12}$ is the count after 11 hours, so the count first exceeds one million after 11 hours.

5. Geometric decay

200 mg of a medicine is in the body, and each hour 30% of it is removed. Find the amount after 5 hours and when it first falls below 10 mg.

Model: 70% remains each hour: $r = 0.7$ . The amount after $k$ hours is $200 \times 0.7^k$ (this is $t_{k+1}$ with $t_1 = 200$ ).
After 5 hours: $200 \times 0.7^5 = 200 \times 0.16807 \approx 33.6$ mg.
Below 10 mg: $200 \times 0.7^8 \approx 11.5$ mg and $200 \times 0.7^9 \approx 8.1$ mg: after 9 hours.

6. Comparing linear and geometric growth

Using the salary plans in the graph, find the first year in which Plan B pays more than Plan A.

Rules: A: $t_n = 50\,000 + 2500(n - 1)$ . B: $t_n = 50\,000 \times 1.04^{n-1}$ .
Trial: Year 12: A $77 500, B $76 972.70 (A ahead). Year 13: A $80 000, B $80 051.61 (B ahead).
Answer: Year 13. Over a longer period, a constant percentage increase always overtakes a constant amount, because the amount added under B keeps growing.

Common traps

The off-by-one error: If $t_1$ is the value now, the value after 6 years is $t_7$ , not $t_6$ . Decide what $t_1$ represents before substituting.
Confusing the rate with the ratio: A 30% decrease gives $r = 0.7$ , not $0.3$ ; a 4% increase gives $r = 1.04$ , not $0.04$ .
Using $n$ instead of $n - 1$: The rules are $t_1 + (n - 1)d$ and $t_1r^{n-1}$ .
Testing only one pair of terms: Check at least two differences (or two ratios) before deciding a sequence is arithmetic (or geometric).
Joining the points: Sequence models are discrete; plot separate points.
Stopping at the wrong term in a "first exceeds" question: Show the terms either side of the target and choose the first one that meets it.
Treating compound growth as linear: Percentage growth is geometric, even if the first few increases look similar.

Exam technique

Start by stating the model type and why ("the same amount is added each year, so arithmetic" or "the same percentage, so geometric"). Write $t_1$ and $d$ or $r$ , then the recursive rule and the rule for $t_n$ . Say which term you need and why (for example "after 6 hours is $t_7$ "). For "when" questions, show the two terms either side of the target. Finish with a sentence in context, and for predictions far into the future, comment on whether the pattern is likely to continue.

Note

Some things go up by the same amount each time, like a taxi meter adding the same price every kilometre: that is an arithmetic sequence, and its graph is a straight line of dots. Other things go up by the same percentage each time, like bacteria doubling every hour: that is a geometric sequence, and its graph curves upward faster and faster. Percentage growth always wins in the end, even if it starts slower, because each increase is bigger than the last.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

The number of songs on a person's playlist, n, in each week since joining a music streaming service, t, forms an arithmetic sequence, as shown by the graph. The graph passes through (1, 87), (2, 209), (3, 331), (4, 453) and (5, 575). Use the arithmetic sequence to predict the number of songs on this person's playlist 25 weeks after joining the streaming service.

Show worked answer →

Step 1: first term (1 mark): $t_1 = 87$ (week 1).
Step 2: common difference (1 mark): $d = 209 - 87 = 122$ (and $331 - 209 = 122$ confirms it).
Step 3: rule (1 mark): $t_n = t_1 + (n - 1)d = 87 + 122(n - 1)$ .
Step 4: week 25 (1 mark): $t_{25} = 87 + 122 \times 24 = 87 + 2928 = 3015$ songs.

The trap is using $n = 24$ , or forgetting the $(n - 1)$ . Week 1 is the first term, so week 25 is the 25th term.

Practice7 marks

The first three lines in a pattern have the equations given. Their slopes form the terms of one sequence and their y-intercepts form the terms of another sequence. Each sequence is either arithmetic or geometric. Line 1: y = -0.8x + 1.2. Line 2: y = 0.4x + 2.7. Line 3: y = -0.2x + 4.2. Determine the coordinates of the point where Line 5 in the pattern intersects Line 1.

Show worked answer →

Step 1: slopes (1 mark): $-0.8, 0.4, -0.2$ is geometric: $r = \frac{0.4}{-0.8} = -0.5$ (and $\frac{-0.2}{0.4} = -0.5$ ), so $m_n = -0.8 \times (-0.5)^{n-1}$ .
Step 2: intercepts (1 mark): $1.2, 2.7, 4.2$ is arithmetic with $d = 1.5$ , so $c_n = 1.2 + 1.5(n - 1)$ .
Step 3: slope of Line 5 (1 mark): $m_5 = -0.8 \times (-0.5)^4 = -0.8 \times 0.0625 = -0.05$ .
Step 4: intercept of Line 5 (1 mark): $c_5 = 1.2 + 1.5 \times 4 = 7.2$ , so Line 5 is $y = -0.05x + 7.2$ .
Step 5: solve (1 mark): $-0.8x + 1.2 = -0.05x + 7.2$ gives $-0.75x = 6$ , so $x = -8$ .
Step 6: $y$ (1 mark): $y = -0.8 \times (-8) + 1.2 = 7.6$ .
Step 7: communicate (1 mark): Line 5 meets Line 1 at $(-8, 7.6)$ .

Practice6 marks

The table shows the total number of times a new song is played on a music service in the days following its first release. Number of days since first release: 5, 10, 15, 20. Total number of times played ('000s): 8, 12, 18, 27. The songwriter is paid 0.175 cents every time their song is played and will be paid after 60 days. They predict that by then they will be owed at least $1000. Given that the number of times the song is played is increasing exponentially, evaluate the reasonableness of this prediction.

Show worked answer →

Under the 2025 syllabus this is a geometric sequence problem (exponential regression is not required).

Step 1: identify the sequence (1 mark). Every 5 days the total is multiplied by the same ratio: $\frac{12}{8} = \frac{18}{12} = \frac{27}{18} = 1.5$ . So it is geometric with $t_1 = 8$ (day 5) and $r = 1.5$ .

Step 2: which term is day 60? (1 mark) Day 5 is $t_1$ , day 10 is $t_2$ , so day 60 is $t_{12}$ .

Step 3: predict (1 mark): $t_{12} = 8 \times 1.5^{11} \approx 691.98$ thousand, about 691 980 plays.
Step 4: pay (1 mark): $691\,980 \times 0.175 = 121\,096.5$ cents, about $1210.97.
Step 5: compare (1 mark): $1210.97 is more than $1000.
Step 6: evaluate (1 mark): The prediction is reasonable if the growth continues. But the model extrapolates from 20 days of data to 60 days, and a song's plays usually level off, so the true amount could be much lower.

Original4 marks

A theatre has 18 seats in row 1, and each row has 3 more seats than the row in front. (a) Write a recursive rule and a rule for

t_n

, the number of seats in row

n

. (b) How many seats are in row 25? (c) Which is the first row with more than 80 seats?

Show worked answer →

(a) Recursive: $t_{n+1} = t_n + 3$ , $t_1 = 18$ . Rule: $t_n = 18 + 3(n - 1)$ . (1 mark)

(b) $t_{25} = 18 + 3 \times 24 = 90$ seats. (1 mark)

(c) $t_{21} = 18 + 3 \times 20 = 78$ and $t_{22} = 18 + 3 \times 21 = 81$ . (1 mark) Row 22 is the first with more than 80 seats. (1 mark)

Original4 marks

A town has 12 000 residents, and its population is falling by 4% each year. (a) Write a recursive rule for the population, with

t_1 = 12\,000

now. (b) Find the population after 10 years. (c) After how many whole years does the population first fall below 8000?

Show worked answer →

(a) Each year 96% remains, so $r = 0.96$ : $t_{n+1} = 0.96t_n$ , $t_1 = 12\,000$ . (1 mark)

(b) After 10 years is $t_{11} = 12\,000 \times 0.96^{10} \approx 7978$ residents. (1 mark)

(c) After 9 years: $12\,000 \times 0.96^9 \approx 8310$ . After 10 years: about 7978. (1 mark) The population first falls below 8000 after 10 years. (1 mark)

Original4 marks

Two salary plans start at $50 000 in year 1. Plan A increases by $2500 each year. Plan B increases by 4% each year. (a) Write the rule for

t_n

for each plan. (b) Find each salary in year 10. (c) In which year does Plan B first pay more than Plan A?

Show worked answer →

(a) A: $t_n = 50\,000 + 2500(n - 1)$ (arithmetic). B: $t_n = 50\,000 \times 1.04^{n-1}$ (geometric). (1 mark)

(b) A: $50\,000 + 2500 \times 9 = 72\,500$ . B: $50\,000 \times 1.04^9 \approx 71\,165.59$ . (1 mark)

(c) Year 12: A $77 500, B about $76 972.70. Year 13: A $80 000, B about $80 051.61. (1 mark) Plan B first pays more in year 13. Geometric growth starts slower here but always overtakes linear growth eventually. (1 mark)

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marks

Classify each sequence as arithmetic, geometric or neither, giving

d

r

: (a) 5, 9, 13, 17 (b) 3, 6, 12, 24 (c) 2, 5, 10, 17.

Show worked solution →

(a) Arithmetic, $d = 4$ . (b) Geometric, $r = 2$ . (c) Neither: the differences are 3, 5, 7 and the ratios are not equal. (2 marks)

foundation2 marks

Generate the first four terms of (a)

t_{n+1} = t_n - 6

t_1 = 40

and (b)

t_{n+1} = 1.5t_n

t_1 = 16

Show worked solution →

(a) 40, 34, 28, 22. (1 mark)

(b) 16, 24, 36, 54. (1 mark)

foundation2 marks

Find

t_{20}

for the arithmetic sequence with

t_1 = 7

and

d = 5

, and

t_8

for the geometric sequence with

t_1 = 3

and

r = 2

Show worked solution →

$t_{20} = 7 + 19 \times 5 = 102$ . (1 mark)

$t_8 = 3 \times 2^7 = 384$ . (1 mark)

core3 marks

A taxi fare for a 1 km trip is $6.30 and each extra kilometre adds $2.10. (a) Write the rule for the fare for an

n

km trip. (b) Find the fare for 18 km. (c) What is the longest whole-number trip that costs no more than $50?

Show worked solution →

(a) $t_n = 6.30 + 2.10(n - 1)$ . (1 mark)

(b) $t_{18} = 6.30 + 2.10 \times 17 = 42.00$ , so $42.00. (1 mark)

(c) $t_{21} = 48.30$ and $t_{22} = 50.40$ , so the longest trip is 21 km. (1 mark)

core3 marks

In an arithmetic sequence

t_3 = 17

and

t_8 = 42

. In a geometric sequence with positive terms,

t_2 = 12

and

t_4 = 48

. Find the first term and the common difference or ratio of each.

Show worked solution →

Arithmetic: from $t_3$ to $t_8$ is 5 steps, so $5d = 42 - 17 = 25$ , $d = 5$ , and $t_1 = 17 - 2 \times 5 = 7$ . (1 mark)

Geometric: from $t_2$ to $t_4$ is 2 steps, so $r^2 = \frac{48}{12} = 4$ , $r = 2$ (positive terms), and $t_1 = \frac{12}{2} = 6$ . (2 marks)

exam3 marks

A dose of 200 mg of a medicine is given, and each hour the amount in the body falls by 30%. (a) Find the amount after 5 hours. (b) After how many whole hours is the amount first below 10 mg?

Show worked solution →

(a) 70% remains each hour: $200 \times 0.7^5 \approx 33.6$ mg. (1 mark)

(b) $200 \times 0.7^8 \approx 11.5$ mg and $200 \times 0.7^9 \approx 8.1$ mg. (1 mark) The amount is first below 10 mg after 9 hours. (1 mark)

exam3 marks

$5000 is invested at 4.5% p.a. simple interest. (a) Write a rule for the balance

t_n

at the end of year

n

. (b) Find the balance after 8 years. (c) In which year does the balance first exceed $7000?

Show worked solution →

(a) Interest per year $= 0.045 \times 5000 = 225$ , so $t_1 = 5225$ , $d = 225$ : $t_n = 5225 + 225(n - 1)$ . (1 mark)

(b) $t_8 = 5225 + 225 \times 7 = 6800$ , so $6800. (1 mark)

(c) $t_8 = 6800$ , $t_9 = 7025$ : the balance first exceeds $7000 at the end of year 9. (1 mark)

Practise this

Sources & how we know this

General Mathematics General senior syllabus 2025 — QCAA